[tex] x^{2} +4 y^{2} -6x+4y+10 \leq 0 <=> \\ <=>( x^{2} -6x+9)+( 4y^{2}+4y+1) \leq 0<=> \\ <=>(x-3)^{2}+(2y+1)^{2} \leq 0[/tex].........(1)
Cum [tex] (x-3)^{2} \geq 0[/tex] si [tex] (2y+1)^{2} \geq 0[/tex], rezulta ca [tex] (x-3)^{2} +(2y+1)^{2} \geq 0 ........(2) [/tex].
Din relatiile (1) si (2) => [tex] (x-3)^{2} +(2y+1)^{2} =0[/tex],dar [tex](x-3)^{2} \geq 0[/tex] si [tex] (2y+1)^{2} \geq 0[/tex] => [tex] (x-3)^{2}=0 => x-3=0 =>x=3 [/tex] si [tex](2y+1)^{2} =0 =>2y+1=0 =>2y=-1=>y=- \frac{1}{2} [/tex].
Solutie: [tex](x;y)=(3; - \frac{1}{2}) [/tex].
